For some dang reason I can't figure out this inequality problem..it's just not working! The solution is $\displaystyle c < -2 $ but I keep getting $\displaystyle c > -2$
Problem: $\displaystyle |\frac{4+c} {c}|<1$
(absolute value of the fraction)
For some dang reason I can't figure out this inequality problem..it's just not working! The solution is $\displaystyle c < -2 $ but I keep getting $\displaystyle c > -2$
Problem: $\displaystyle |\frac{4+c} {c}|<1$
(absolute value of the fraction)
$\displaystyle |\frac{4+c} {c}|<1\Longleftrightarrow -1<\frac{4+c}{c}< 1$ , so the next inequalities have both to be true:
1) $\displaystyle -1< \frac{4+c}{c}\Longrightarrow 2\,\frac{2+c}{c}> 0\Longrightarrow c< -2\,\,\,or\,\,\,c> 0$
2) $\displaystyle \frac{4+c}{c}< 1\Longrightarrow \frac{4}{c}< 0\Longrightarrow c< 0$
From (1)(2) we get $\displaystyle c< -2$
Tonio
Hi Tonio--I still don't understand how you got line (1) and wouldn't line (2) produce $\displaystyle 4 < 0 $ and not $\displaystyle c < 0 $ the former being nonsensical since the variable is gone when multiplied by 0.
Here's why I come out with $\displaystyle c > -2 $ :
$\displaystyle |\frac{4+c} {c}|<1$
$\displaystyle \frac{4}{c} + 1>-1$
$\displaystyle \frac{4}{c}>-2$
$\displaystyle (c)\frac{4}{c}>-2c => 4>-2c$
Flip signs since dividing by a negative! $\displaystyle -2<c => c>-2$
1)
$\displaystyle -1 < \frac{4+c}{c} $
$\displaystyle 0 < \frac{4+c}{c} +1$
$\displaystyle 0< \frac{4+2c}{c}$
now set your denominator and numerator equal to 0
and you get c equals -2 and 0
Now test a value to the left of -2 on the number line, test a value between -2 and 0, and then test a value to the right of 0 into the inequality
$\displaystyle 0< \frac{4+2(-3)}{-3}= \frac{2}{3}$
which makes the inequality true
$\displaystyle 0< \frac{4+2(-1)}{-1} = -2$
which makes the inequality false
$\displaystyle 0< \frac{4+2(1)}{1)} = 6$
Which makes the inequality true
so $\displaystyle c < -2$ or $\displaystyle c>0$
2)
$\displaystyle 1> \frac{4+c}{c}$
$\displaystyle 0> \frac{4+c}{c} -1$
$\displaystyle 0> \frac{4+c}{c} -\frac{c}{c}$
$\displaystyle 0> \frac{4}{c}$
set your numerator and denominator equal to 0 again and you get c = 0
Now test a value to the left of o and to the right of zero
$\displaystyle 0> \frac{4}{-1}= -4$
which is true
$\displaystyle 0> \frac{4}{1}= 4$
which is false.
so
c<0
Now putting all this you notice the inequality of
$\displaystyle c<0$ and $\displaystyle c<-2$ overlap so
$\displaystyle c<-2$
is your answer.